HU Berlin Winter Term 2012/13 Exercises Algebra II (Commutative Algebra) Prof. Dr. J. Kramer To be handed in on November 5th after the lecture Please hand in every exercise solution on a seperate sheet and do not forget to put your name and student ID on every sheet. Exercise sheet 3 (50 points) Exercise 1 (10 points) Let A be a commutative ring with 1. Prove the following assertions: (a) If every x 2 A satisfies xn = x for some integer n ≥ 2, then show that every prime ideal in A is a maximal ideal. (b) An element x 2 A is called an idempotent, if x2 = x. Show that a local ring A has no idempotents different from 0 and 1. Exercise 2 (10 points) Let A be a commutative ring with 1. Let A[X] be the ring of polynomials in X with n coefficients in A and let f = anX +:::+a1X +a0 2 A[X]. Prove the following assertions: (a) f is a unit in A[X] if and only if a0 is a unit in A and a1; : : : ; an are nilpotent. (b) f is nilpotent if and only if a0; a1; : : : ; an are nilpotent. (c) f is a zero-divisor if and only if there exists a 2 A, a 6= 0, such that af = 0. Exercise 3 (10 points) Prove the following assertions: (a) Show that the Z-module Q is not free. Furthermore, prove that Z=nZ is not a free Z-module, but a free Z=nZ-module. (b) Let A be a commutative ring with 1. Show that, if every finitely generated A-module M is free, then A = f0g or A is a field. (c) Find a commutative ring A with 1, an A-module M, and a set of generators fx1; : : : ; xng of M, such that fx1; : : : ; xng is not minimal, but such that fx1; : : : ; xn−1g does not generate M anymore. Please turn over ! Exercise 4 (10 points) Let A be a commutative ring with 1. An ideal a ( A is called an idempotent ideal if a2 = a. Let a ( A be a finitely generated idempotent ideal of A. Then, show that there exists an element e 2 a such that e2 = e and a = eA (i.e., a is a principal ideal generated by the element e). Exercise 5 (10 points) Let A be a commutative ring with 1. For f 2 A, we define the distinguished or basic set D(f) := Spec(A) n V (f) to be the complement of V (f) in Spec(A). Show that the sets D(f) (f 2 A) are open and that they form a basis of open sets for the Zariski topology of Spec(A). Furthermore, prove that for f; g 2 A we have: (a) D(f) \ D(g) = D(f · g). (b) D(f) = ; () f is nilpotent. (c) D(f) = X () f 2 A×. (d) D(f) = D(g) () r(f) = r(g): Here, r(f) denotes the radical of the principal ideal (f)..